Find 3 consecutive integers such that the product of the middle and largest integers is five more then the smallest integer squared

Only option: 1,2,3

Step-by-step explanation:

3 consecutive integers:

n-1, n, n+1

Why does it work? They should be consecutive for any integer n.

Example with n=5:

4, 5, 6

These are consecutive integers because they happen right after each other in the list of integers.

Integers { ..., - 5,-4,-3,-2,-1,0,1,2,3,4,5, ... }

We are given the product of the middle and largest is 5 more than the smallest integer squared.

In other words, the product of n and (n+1) is 5 more than (n-1) ^2.

As an equation that is: n (n+1) = 5 + (n-1) ^2.

Let's write it more together:

n (n+1) = 5 + (n-1) ^2

Distribute:

n^2+n=5+n^2-2n+1 (Used (x-y) ^2=x^2-2xy+y^2)

Combine like terms on either side:

n^2+n=n^2-2n+6

Get everything on one side so one side is 0.

Subtract n^2 on both sides:

n=-2n+6

Add 2n on both sides:

3n=6

Divide both sides by 3:

n=2

So if n=2,

then n-1=1

and n+1=3.

So we have one option (1,2,3).

3 (2) = 5 + (1) ^2

6=5+1

6=6 is true so (1,2,3) is an option.